Question

The area bounded by the curve \[y = x{\left( {3 - x} \right)^2}\], the \[x - axis\] and the ordinates of the maximum and minimum points of the curve is
A. $2$ sq. unit
B. $4$ sq. unit
C. $6$ sq. unit
D. $8$ sq. unit

Answer 1

Hint: : In this question, we are given the equation of curve and we have to find the area bounded by the maximum and minimum points of the curve. For minimum and maximum points, differentiate the given equation of curve using the chain rule and equate it to zero. Then, plot the graph of the required coordinates. In last, to calculate the area integrate the equation of curve with respect to $dx$ from $x = 1$ to $x = 3$ and solve it further using the integration formula.



Formula Used:Chain rule of differentiation –
$\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$
Algebraic identity –
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Integration formula –
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $



Complete step by step solution:Given that,
Equation of the curve is \[y = x{\left( {3 - x} \right)^2}\]
Differentiate the given equation of curve \[y = x{\left( {3 - x} \right)^2}\] with respect to $x$,
Applying the chain rule of differentiation $\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$
We get, $\dfrac{{dy}}{{dx}} = x\left( {2\left( {3 - x} \right)\left( { - 1} \right)} \right) + {\left( {3 - x} \right)^2}$
Using algebraic identity, ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
On solving, we get $\dfrac{{dy}}{{dx}} = {x^2} - 4x + 3$
For maximum and minimum points on the curve take $\dfrac{{dy}}{{dx}}$ equal to $0$,
It implies that,
${x^2} - 4x + 3 = 0$
Using middle term split factorization, we get $\left( {x - 1} \right)\left( {x - 3} \right) = 0$
We get, $x = 1$ and $x = 3$
Put the required values in the equation of curve.
Thus, the maximum and the minimum points on the curve are $\left( {1,4} \right)$ and $\left( {3,0} \right)$.
Graph of the given equations and the required coordinates is attached below (figure 1);

Figure 1: A Graph contains the plot of the equation of the curve and its maximum and minimum points
Now, to calculate the area of the bounded region integrate the equation of the curve with respect to $dx$ from $x = 1$ to $x = 3$
Therefore, the required area of the bounded region will be
$A = \int\limits_1^3 {x{{\left( {3 - x} \right)}^2}dx} $
Here, $x{\left( {3 - x} \right)^2} = x\left( {9 + {x^2} - 6x} \right) = 9x + {x^3} - 6{x^2}$
Thus, area $A = \int\limits_1^3 {\left( {9x + {x^3} - 6{x^2}} \right)dx} $
As we know that, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
It implies that, $A = \left[ {\dfrac{{9{x^2}}}{2} + \dfrac{{{x^4}}}{4} - \dfrac{{6{x^3}}}{3}} \right]_1^3$
On resolving the limits, we get $A = \dfrac{{9{{\left( 3 \right)}^2}}}{2} + \dfrac{{{{\left( 3 \right)}^4}}}{4} - \dfrac{{6{{\left( 3 \right)}^3}}}{3} - \left( {\dfrac{{9{{\left( 1 \right)}^2}}}{2} + \dfrac{{{{\left( 1 \right)}^4}}}{4} - \dfrac{{6{{\left( 1 \right)}^3}}}{3}} \right)$
On solving, we get $A = 4$ sq. units



Option ‘B’ is correct

Note: Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.